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Thermodynamics - Calorific Capacity

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a system with a constant number of particles.
    Write the total differential dS in terms of the derivarives [itex]\frac{∂S}{∂T}[/itex] and [itex]\frac{∂S}{∂V}[/itex]. Introduce CV (calorific capacity at constant volume).
    Next write the total differential of the volume dV in terms of the parcial derivatives [itex]\frac{∂V}{∂T}[/itex] and [itex]\frac{∂V}{∂P}[/itex]. Assume that the pressure is constant. Show that the result comes in the form of:

    CP-CV= Expression

    2. Relevant equations

    CP=T[itex]\frac{∂S}{∂T}[/itex] , P and N Constant
    CV=T[itex]\frac{∂S}{∂T}[/itex] , V and N Constant


    3. The attempt at a solution

    First I wrote the differential of the entropy as asked:

    dS = [itex]\frac{∂S}{∂T}[/itex]dT + [itex]\frac{∂S}{∂V}[/itex]dV

    I know that [itex]\frac{∂S}{∂V}[/itex] = CV/T. Substituting I get:

    dS = CV/T dT + [itex]\frac{∂S}{∂V}[/itex]dV

    Next I found the differential of the volume:

    dV = [itex]\frac{∂V}{∂T}[/itex]dT + [itex]\frac{∂V}{∂P}[/itex]dP

    Since the pressure is constant it reduces to the form

    dV = [itex]\frac{∂V}{∂T}[/itex]dT

    Substituting in our dS expression we get:

    dS = CV/T dT + [itex]\frac{∂S}{∂V}[/itex][itex]\frac{∂V}{∂T}[/itex]dT =
    = CV/T dT + [itex]\frac{∂S}{∂T}[/itex]dT =
    = CV/T dT + CP/T dT ⇔
    ⇔ T[itex]\frac{dS}{dT}[/itex] = CV + CP

    I'm making some mistake. If anyone could point me in the right direction I'd appreciate.

    Thanks!
     
  2. jcsd
  3. Nov 25, 2012 #2

    haruspex

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    I think your problem is a confusion over the meanings of different partial derivatives. Your use of ##\frac{∂S}{∂V}## refers to changing volume, keeping temperature constant.
    That's inconsistent with later equating ##\frac{∂S}{∂V}\frac{∂V}{∂T} = \frac{∂S}{∂T}##, which, to be valid, assumes pressure constant in all terms.
     
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